Congruum Problem

Date: 04/04/2002 at 18:06:35
From: Dr. Allan
Subject: Congruum problem
In researching suitable problems for my students I have found a
reference to Fibonacci and his congruum problem. Something has me
stumped; I hope one or more of you can tell me what I am doing wrong:
Quote from the article at the MacTutor History of Mathematics archive
at St. Andrews University:
http://www-history.mcs.st-and.ac.uk/history/Mathematicians/Fibonacci.html
"[Fibonacci] defined the concept of a congruum, a number of the form
ab(a + b)(a - b), if a + b is even, and 4 times this if a + b is odd
where a and b are integers. Fibonacci proved that a congruum must be
divisible by 24 and he also showed that for x,c such that x^2 + c and
x^2 - c are both squares, then c is a congruum. He also proved that a
square cannot be a congruum."
With x=15 and c=216, we get the two squares 441 and 9, meaning that
216 should be a congruum. Thus we should be able to find numbers a
and b such that
216 = ab(a+b)(a-b) if a+b is even
or
54 = ab(a+b)(a-b) if a+b is odd
Now, b must be smaller than a and d:= ab(a+b)(a-b) is increasing in
both a and b.
Let's look at the possible scenarios.
Say a+b is odd. We want d to equal 54. Our triples (b,a,d) give us
(1,2,6) (1,4,60); thus b=1 is not possible
(2,3,30) (2,5,210); thus b=2 is not possible
(3,4,84); thus b = 3 and b > 3 is not possible (since d is
increasing); thus a+b odd is not possible
Say a+b is even. We want d to equal 216. Our triples (b,a,d) give us
(1,5,120) (1,7,336); thus b=1 is not possible
(2,4,96) (2,6,3849; thus b=2 is not possible
(3,5,360); thus b = 3 and b > 3 is not possible (since d is
increasing); thus a+b even is not possible
So, because 15^2 + 216 = 21^2 and 15^2 - 216 = 3^2 we should have
that 216 is a congruum, but we cannot put it in the form ab(a+b)(a-b).
What is wrong with this argument?
Sincerely,
Dr. Allan

Date: 04/05/2002 at 16:04:44
From: Dr. Paul
Subject: Congruum problem
I went to the library this morning and picked up a copy of Ore's
_Number Theory and its History_. The solution to the congruum problem
is listed on pages 188-193 and I found it very interesting reading.
What follows is essentially Ore's solution with my commentary added
where I felt Ore's details were (probably intentionally) lacking
sufficient explanation:
We want to find a number x such that simultaneously:
x^2 + h = a^2, x^2 - h = b^2 (equation 8-16)
and determine for which h rational solutions x can exist. We shall
first determine the solutions in integers, and this depends, as we
shall see again, on the Pythagorean triangle. When the second equation
in (8-16) above is subtracted from the first, one has:
2*h = (a^2 - b^2) = (a-b)*(a+b) (8-17)
Since the left-hand side is even, a and b must both be odd or both be
even. Therefore, a - b is even. Hence
a - b = 2*k
and k must be a divisor of h since according to (8-17) we have:
2*h = 2*k*(a+b)
or
h = k*(a+b)
which implies k divides h.
It follows that
(a + b) = h/k
and by adding and subtracting the last two equations, one finds
a = h/(2*k) + k, b = h/(2*k) - k
When these two expressions are substituted into the original equations
(8-16), we obtain:
x^2 + h = [h/(2*k) + k]^2 = [h/(2*k)]^2 + h + k^2
and
x^2 - h = [h/(2*k) - k]^2 = [h/(2*k)]^2 - h + k^2
Adding these equations and removing a common factor of two gives:
x^2 = [h/(2*k)]^2 + k^2
In our case of interest, we have:
x = 15
h = 216
a = 21
b = 3
k = 9
Therefore, the three numbers
x, h/(2*k), k
form a Pythagorean triangle. Determine the lesser of h/(2*k) and k (in
this case k = 9 < 12 = 216/18 = h/(2*k)) so that we can write:
x = t*(m^2 + n^2)
k = t*(m^2 - n^2) (**)
h/(2*k) = 2*m*n*t
where t is some integer and the expressions in m and n define a
primitive solution of the triangle.
In actuality, we have:
x = 15
k = 9
h = 216
so clearly, t = 3 and this forces m = 2, n = 1
As an aside, if h/(2*k) > k then we pick
h/(2*k) = t*(m^2 - n^2)
k = 2*m*n*t
Now we take the product of the last two expressions in (**) above to
obtain as the general solution to (8-16)
x = t*(m^2 + n^2), h = 4*m*n*(m^2 - n^2)*t^2 (8-18)
Now we make a slight reduction to this solution. Let us suppose that
we have a solution x of (8-16), where x has the factor t and h at the
same time has the factor t^2 (this is the case in our example since
x = 15 is divisible by three and h = 216 is divisible by 9). Then we
can write:
x = x_1 * t, h = h_1 * t^2
From (8-16), we obtain:
[(x_1)^2 * t^2] + (h_1 * t^2) = a^2
and
[(x_1)^2 * t^2] - (h_1 * t^2) = b^2
Then a^2 and b^2 are both divisible by t^2 so a and b are both
divisible by t:
a = a_1 * t, b = b_1 * t
This gives:
[(x_1)^2 * t^2] + (h_1 * t^2) = [(a_1)^2 * t^2]
and
[(x_1)^2 * t^2] - (h_1 * t^2) = [(b_1)^2 * t^2].
Cancelling the factor of t^2 gives:
(x_1)^2 + h_1 = (a_1)^2
and
(x_1)^2 - h_1 = (b_1)^2
When no further reduction is possible, we shall say that we have a
primitive solution. When this reduction is applied to (8-18), we
obtain:
x = m^2 + n^2
h = 4*m*n*(m^2 - n^2)
The numbers m and n now produce a primitive Pythagorean triple.
Thus your solution of
15^2 + 216 = 21^2
15^2 - 216 = 3^2
Is really just a multiple of the primitive solution:
5^2 + 24 = 7^2
5^2 - 24 = 1^2
This corresponds to the first row in the table here:
Congruum Problem - MathWorld - Eric Weisstein
http://mathworld.wolfram.com/CongruumProblem.html
i.e., take m = 2, n = 1. Then x = 5, h = 24, a = 7, b = 1
and as with Pythagorean triples, letting t be a natural number gives
an infinite number of solutions.
Neat problem!
I hope this is clear - let me know if I'm skipping steps where I need
to be filling in the details.
Finally, notice that writing
h = 4*m*n*(m-n)*(m+n)
is only possible when h is an element of a primitive congruum. That's
why your attempts failed. You needed to write
h = 4*m*n*(m-n)*(m+n)*t^2
which works easily when h = 216, m = 2, n = 1, t = 3.
- Doctor Paul, The Math Forum
http://mathforum.org/dr.math/